## Plane Wave Solutions of the Dirac Equation

As usual, the Dirac equation can be solved by Fourier decomposition in plane waves,

(x) = ∫

d3 k√ 2k0(k)(2π )3

˜(k)e−ikx. (13.1)

where ˜(k) are complex four-vectors that satisfy

(k/ − m)˜(k) ≡ (kµγ µ − m)˜(k) = 0, k20 = k 2 + m2. (13.2) The Lorentz invariance implies

˜ ′(k ′) = S(ω)˜(k), k ′µ = K (ω)µνkν . (13.3)

As any k can be obtained from a boost to k = 0, all solutions of the Dirac equation can be obtained from the ones at rest with k = 0 (see Problem 19)

(γ 0k0 − m)˜(0) = (k0 − m)12

−(k0 + m)12

˜(0). (13.4)

We see that for k0 > 0 (k0 = m), there are two independent solutions both of the form

˜+(0) = A

, (13.5)

where A is a spin one-half two-spinor of which

1

is the spin-up and

0

is the spin-down state. The identification of the spin degrees of freedom

follows from the behaviour of ˜ under rotations (which leave k = 0). We leave it as an exercise to verify that also in the Dirac representation of the gamma

A in

matrices, like in the Weyl representation [see Equation (12.25)],

σi j = εi jk σk

σk

, (13.6)

such that A is easily seen to transform under a rotation as in Eq. (12.8) [compare this to the discussion following Equation (12.25)]. There are also two solutions for k0 < 0 (k0 = −m) of the form

˜−(0) = 0

, (13.7)

where likewise B is a spin one-half two-spinor of which

1

is the spin-up

and

0

is the spin-down state, which transform under rotations as A.